What is the maximum value of this nested radical?

I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:

What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?

Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:

What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.

Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.

I’d appreciate proofs to my thoughts, especially the last one!

functions recursion maxima-minima nested-radicals golden-ratio

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edited Jan 5 at 19:26

TheSimpliFire

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

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This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.

The current answers do not contain enough detail.

This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.

1

$begingroup$How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.$endgroup$
– T_MJan 5 at 17:20

add a comment |

22

8

$begingroup$

I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:

What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?

Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:

What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.

Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.

I’d appreciate proofs to my thoughts, especially the last one!

functions recursion maxima-minima nested-radicals golden-ratio

share|cite|improve this question

edited Jan 5 at 19:26

TheSimpliFire

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

$endgroup$

This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.

The current answers do not contain enough detail.

This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.

1

$begingroup$How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.$endgroup$
– T_MJan 5 at 17:20

add a comment |

22

8

22

8

22

8

$begingroup$

I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:

What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?

Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:

What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.

Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.

I’d appreciate proofs to my thoughts, especially the last one!

functions recursion maxima-minima nested-radicals golden-ratio

share|cite|improve this question

edited Jan 5 at 19:26

TheSimpliFire

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

$endgroup$

I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:

What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?

Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+ sqrt{ frac{x^4}{x-sqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:

What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.

Furthermore, I also conjecture that $$max F_infty=phi-1=frac{sqrt5-1}2,$$ the positive solution of the equation $x^2+x-1=0$. This seems right as $max F_{15}=0.6179$ from the plot.

I’d appreciate proofs to my thoughts, especially the last one!

functions recursion maxima-minima nested-radicals golden-ratio

functions recursion maxima-minima nested-radicals golden-ratio

share|cite|improve this question

edited Jan 5 at 19:26

TheSimpliFire

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

share|cite|improve this question

edited Jan 5 at 19:26

TheSimpliFire

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

share|cite|improve this question

share|cite|improve this question

edited Jan 5 at 19:26

TheSimpliFire

edited Jan 5 at 19:26

TheSimpliFire

edited Jan 5 at 19:26

TheSimpliFire

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

asked Jan 5 at 16:41

TheSimpliFireTheSimpliFire

12.6k62360

12.6k62360

This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.

The current answers do not contain enough detail.

This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.

This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.

The current answers do not contain enough detail.

This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.

1

$begingroup$How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.$endgroup$
– T_MJan 5 at 17:20

add a comment |

1

$begingroup$How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.$endgroup$
– T_MJan 5 at 17:20

1

1

$begingroup$How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.$endgroup$
– T_MJan 5 at 17:20

$begingroup$How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.$endgroup$
– T_MJan 5 at 17:20

add a comment |

1 Answer1

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oldest

votes

12

$begingroup$

Zachary’s and OP’s combined solutions

If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:

$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.

Proof attempt for the domain:

While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$$exists y mid forall n geq y, F_n(x) notin mathbb{R}$.

Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.

Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.

Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$. Since $x$ is non-negative, the domain is $[0,1]$.

So, for any $x$ there’s an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.

Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.

As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go non-real. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.

Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)

Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)-x(1+G’)}{(x+G)^2}>0impliedby G-xG’>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.

Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H’x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H’x^2-H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.

share|cite|improve this answer

edited 2 days ago

TheSimpliFire

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answered Jan 5 at 17:37

Zachary HunterZachary Hunter

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$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

1

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

Post as a guest

1 Answer1

1 Answer1

If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:

$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.

Proof attempt for the domain:

While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$$exists y mid forall n geq y, F_n(x) notin mathbb{R}$.

Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.

Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.

Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$. Since $x$ is non-negative, the domain is $[0,1]$.

So, for any $x$ there’s an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.

Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.

As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go non-real. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.

Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)

Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)-x(1+G’)}{(x+G)^2}>0impliedby G-xG’>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.

Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H’x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H’x^2-H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.

share|cite|improve this answer

edited 2 days ago

TheSimpliFire

12.6k62360

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

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1

$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

1

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

add a comment |

12

$begingroup$

Zachary’s and OP’s combined solutions

If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:

$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.

Proof attempt for the domain:

While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$$exists y mid forall n geq y, F_n(x) notin mathbb{R}$.

Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.

Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.

Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$. Since $x$ is non-negative, the domain is $[0,1]$.

So, for any $x$ there’s an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.

Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.

As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go non-real. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.

Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)

Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)-x(1+G’)}{(x+G)^2}>0impliedby G-xG’>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.

Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H’x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H’x^2-H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.

share|cite|improve this answer

edited 2 days ago

TheSimpliFire

12.6k62360

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

$endgroup$

1

$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

1

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

add a comment |

12

12

12

$begingroup$

Zachary’s and OP’s combined solutions

If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:

$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.

Proof attempt for the domain:

While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$$exists y mid forall n geq y, F_n(x) notin mathbb{R}$.

Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.

Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.

Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$. Since $x$ is non-negative, the domain is $[0,1]$.

So, for any $x$ there’s an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.

Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.

As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go non-real. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.

Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)

Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)-x(1+G’)}{(x+G)^2}>0impliedby G-xG’>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.

Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H’x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H’x^2-H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.

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edited 2 days ago

TheSimpliFire

12.6k62360

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

$endgroup$

Zachary’s and OP’s combined solutions

If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:

$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1- sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1- F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1- F_infty (1) }} right) = 1$$ so $$(F_infty(1)^2-1)^2=frac{F_infty(1)^4}{1-F_infty(1)}implies F_infty(1)^5-2F_infty(1)^3+2F_infty(1)^2+F_infty(1)-1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)-1)(F_infty(1)^3-F_infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_infty(1)=phi-1$ derived from the first quadratic factor.

Proof attempt for the domain:

While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$$exists y mid forall n geq y, F_n(x) notin mathbb{R}$.

Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.

Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.

Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$. Since $x$ is non-negative, the domain is $[0,1]$.

So, for any $x$ there’s an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.

Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.

As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go non-real. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x-1}$.

Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)

Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{x-sqrt{frac{x^3}{x+sqrt{frac{x^4}{x- sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{x-sqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)-x(1+G’)}{(x+G)^2}>0impliedby G-xG’>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.

Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2-H}{(x-H)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{x-H}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2-H}{(x-H)^2}&impliedbyfrac{2x}{x-H}>1+frac{H’x^2-H}{(x-H)^2}\&impliedby 2x^2-2Hx>x^2-2Hx+H^2+H’x^2-H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.

share|cite|improve this answer

edited 2 days ago

TheSimpliFire

12.6k62360

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

share|cite|improve this answer

share|cite|improve this answer

edited 2 days ago

TheSimpliFire

12.6k62360

edited 2 days ago

TheSimpliFire

12.6k62360

edited 2 days ago

TheSimpliFire

12.6k62360

12.6k62360

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

answered Jan 5 at 17:37

Zachary HunterZachary Hunter

54111

54111

1

$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

1

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

add a comment |

1

$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

1

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

1

1

$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.$endgroup$
– Zachary HunterJan 5 at 18:18

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

$begingroup$Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.$endgroup$
– Zachary HunterJan 7 at 0:43

1

1

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

$begingroup$Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!$endgroup$
– TheSimpliFire2 days ago

add a comment |

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